# How many real numbers satisfy x+x^(−1) so that it is a Whole number. I came across this in a discrete mathematics book, and i wonder if my solution is correct. From looking at y=x+x^(−1) I see only 2 possibilities, x ={1;-1;} Was this just so easy or am I missing something

How many real numbers satisfy $x+{x}^{-1}$ so that it is a Whole number.
I came across this in a discrete mathematics book, and i wonder if my solution is correct.
From looking at $y=x+{x}^{-1}$
I see only 2 possibilities, x ={1;-1;}
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domino671v
For any integer n we want
$n=x+\frac{1}{x}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-nx+1=0$
We need $\mathrm{\Delta }={n}^{2}-4\ge 0$ for the above equation to have solutions, so $|n|\ge 2$
And the solutions are $x=\frac{n±\sqrt{{n}^{2}-4}}{2}$ (note that ${x}^{-1}=\frac{n\mp \sqrt{{n}^{2}-4}}{2}$ in that case).
For $n=2$ you get the $\left(1,1\right)$ solution
For $n=-2$ you get $\left(-1,-1\right)$ solution
For $n=4$ you get the nice $\left(2+\sqrt{3},2-\sqrt{3}\right)$ solution
and so on ...
###### Did you like this example?
Aidyn Crosby
Guide:
Let
$x+\frac{1}{x}=k$
${x}^{2}+1=kx$
This is a quadratic equation. For which values of k does this has a real solution?